INTEGRAL TRANSFORMS
Prove relation (13.23).Considering the integral
∫∞
−∞f(t)H′(t)dt=[
f(t)H(t)]∞
−∞−
∫∞
−∞f′(t)H(t)dt=f(∞)−∫∞
0f′(t)dt=f(∞)−[
f(t)]∞
0=f(0),and comparing it with (13.12) whena= 0 immediately shows thatH′(t)=δ(t).
13.1.4 Relation of theδ-function to Fourier transformsIn the previous section we introduced the Diracδ-function as a way of repre-
senting very sharp narrow pulses, but in no way related it to Fourier transforms.
We now show that theδ-function can equally well be defined in a way that more
naturally relates it to the Fourier transform.
Referring back to the Fourier inversion theorem (13.4), we havef(t)=1
2 π∫∞−∞dω eiωt∫∞−∞du f(u)e−iωu=∫∞−∞du f(u){
1
2 π∫∞−∞eiω(t−u)dω}
.Comparison of this with (13.12) shows that we may write theδ-function as
δ(t−u)=1
2 π∫∞−∞eiω(t−u)dω. (13.24)Considered as a Fourier transform, this representation shows that a verynarrow time peak att=uresults from the superposition of a complete spectrum
of harmonic waves, all frequencies having the same amplitude and all waves being
in phase att=u. This suggests that theδ-function may also be represented as
the limit of the transform of a uniform distribution of unit height as the width
of this distribution becomes infinite.
Consider the rectangular distribution of frequencies shown in figure 13.4(a).From (13.6), taking the inverse Fourier transform,
fΩ(t)=1
√
2 π∫Ω−Ω1 ×eiωtdω=2Ω
√
2 πsin Ωt
Ωt. (13.25)
This function is illustrated in figure 13.4(b) and it is apparent that, for large Ω, it
becomes very large att= 0 and also very narrow aboutt= 0, as we qualitatively